Question

A ball of massfalls downward, as shown in Figure 6.82. Initially you observe it to beabove the ground. After a short time it is just about to hit the ground.

(a) During this interval how much work was done on the ball by the gravitational force? (b) Does the kinetic energy of the ball increase or decrease?

(c) The ball hits the ground and bounces back upward, as shown in Figure 6.83. After a short time it isabove the ground again. During this second interval (between leaving the ground and reaching a height of) how much work was done on the ball by the gravitational force? (d) Does the kinetic energy of the ball increase or decrease?

Short answer

a) , 30.87 J

b) the kinetic energy of the ball will increase,

c) $-30.87J$and

d) the kinetic energy of the ball will decrease.

Step-by-step solution

Identification of the given data

The given data can be listed below as-

• The mass of the ball is,$m=0.7kg$ .
• The initial distance of the ball is, $d_1=4.5m$.
• The final distance of the ball is, $d_2=4.5m$.

Significance of the work done

The work done is the product of the force exerted on an object and the distance through which the force has been applied.

The equation of the work done gives the amount of work done in different intervals that further gives the increase or decrease in the kinetic energy.

Determination of the work done by the gravitational force in the first interval

The equation of the work done by the ball is expressed as follows,

$$\begin{gathered} W=F{d}_1 \\ =mg{d}_1 \end{gathered}$$

Here, $F=mg$is the applied force, (m and g are the mass of the ball and the acceleration due to gravity that is $9.8m/s^2$) and is the initial distance of the ball.

For $m=0.7kg,g=9.8m/s^2$, and $d_1=4.5m$.

$$\begin{gathered} W=\left(0.7\right)\times \left(9.8m/s^2\right)\times \left(4.5m\right) \\ =30.87kg{m}^2/s^2 \\ =30.87kg{m}^2/s^2 \\ =30.87J \end{gathered}$$

Thus, the amount of work done by the gravitational force is 30.87 J.

Determination of the kinetic energy of the ball in the first interval

b)

According to the work-energy theorem, the net work done is equal to the change in the kinetic energy. So, if the work done increases, then there will be increment in the kinetic energy also.

Thus, the kinetic energy of the ball will increase.

Determination of the work done by the gravitational force in the second interval

(c)

From the work-energy theorem, the work done by the ball is expressed as-

$$\begin{gathered} W=F{d}_2 \\ =mg{d}_2 \end{gathered}$$

Here, is the applied force, (m and g are the mass of the ball and the acceleration due to gravity that is as the ball moved against gravity) and is the final distance of the ball